Question

A certain kind of sheet metal has, on average, 7 defects per 11 square feet. Assuming a Poisson distribution, find the probability that a 21 square foot metal sheet has at least 11 defects. Round your answer to four decimals.

Answer 1

`p(x\geq 11)=0.7782`

Explanation:

If a variable x follows a poisson distribution, the probability that occurs x events in a region of size t is:

`p(x)=(e^(-mt)*(mt)^(x))/(x!)`

Where m is the mean per unit. So, replacing m by 7/11 because there are 7 defects per 11 square feet and t by 21 square feet, the probability that the metal sheet has x defects is:

`p(x)=(e^(-(7/11)21)*((7/11)(21))^(x))/(x!)=(e^(-147/11)*(147/11)^(x))/(x!)`

Then, the probability that a 21 square foot metal sheet has at least 11 defects is calcualted as:

`p(x\geq 11)=1-p(x\leq 10)`

Where
`p(x\leq 10)=p(0)+p(1)+p(2)+...+p(8)+p(9)+p(10)`

Now, p(0) is equal to:

`p(0)=(e^(-147/11)*(147/11)^(0))/(0!)=0.00000157`

At the same way we can calculated the other probabilities, so:

`p(x\leq 10)=0.2218\\p(x\geq 11)=1-0.2218=0.7782`